Let $G$ be a group generated by a set $S$.
(1) If $|S| < \aleph_0$, then can it be said that $|G|\leq \aleph_0$ ?
(2) In general if $|S| \leq \kappa$, can it be said that $|G| \leq \kappa$?
My reasoning for the question is that every element $g \in G$ can be written as a finite products of elements from $S$. Then if I denote $G_n$ as the elements in $G$ that can be written as a product of $n$ elements in $S$ , for every $n \in \mathbb{N}$, I get that $G \subseteq \cup_{n=1}^\infty G_n$.
And then I get that: $|G| \leq \sum_{n=1}^\infty |G_n| \leq \sum_{n=1}^\infty |S|$, and if $|S|\geq \aleph_0$ then $|G| \leq |S|$.
I'm not sure about this reasoning because it seems to me too simple a result, and would appreciate any input whether to validity of the argument or to the answer.
It is indeed true, it's a special case of the downward Löwenheim-Skolem theorem.
Your proof is fine, and if you want to detail it to really see that it works, this is the main point you may want to look at : Prove that any $g\in G$ can be written as a finite product of elements of $S$. This is a standard argument, but maybe actually proving it can help you make sure you understand it.
EDIT : Why is this a special case of the Downward Löwenheim-Skolem theorem ? I'll deal with the case $|S| \leq \aleph_0$, I'll let you generalize it. Consider the (finite) language of groups $L := \{\cdot, ^{-1}, e\}$ and the theory $T$ of groups. Then $G$ (with the operations) is an $L$-structure. Moreover $S\subset G$ is of cardinality $\leq \aleph_0$, and as $L$ is finite, Löwenheim-Skolem gives the existence of an elementary substructure $N$ of $G$ containing $S$ of cardinality $\leq \aleph_0$ Such a substructure is a group, and as the smallest subgroup of $G$ containing $S$ is $G$ (by definition of generating set), we must have that $N=G$, but then $|G|= |N| \leq \aleph_0$.